The electric field of cylinder at a distance [tex]r[/tex] more than radius of cylinder is equal to [tex]\boxed{E=\dfrac{\lambda}{2\pi\epsilon_0r}}}[/tex].
Further Explanation:
Gauss law states that net electric flux through any closed surface is equal to [tex]\dfrac{1}{\epsilon_0}[/tex] times the net electric charge within that closed surface. The hypothetical closed surface is name as Gaussian surface. The electric flux also defined as the total electric field lines, which passes through a surface area of any closed curve.
Given:
The linear charge density of cylinder is [tex]\lambda[/tex].
Concept:
Consider the length of the cylinder is [tex]L[/tex].
Charge on curved surface of cylinder is [tex]\lambda L[/tex].
For a Gaussian curve in the shape of a concentric cylinder (with a radius more than the radius of charged cylinder) we can write gauss law as:
[tex]\phi=\dfrac{q}{\epsilon_0}[/tex]
Substitute [tex]\lambda L[/tex] for [tex]q[/tex] in above equation.
[tex]\phi=\dfrac{\lambda L}{\epsilon_0}[/tex] …… (I)
Here, [tex]\phi[/tex] is the Electric flux and [tex]\epsilon_0[/tex] is the permittivity of vacuum.
Since the electric field is constant for a given distance [tex]r[/tex] from the axis of the cylinder we can write that:
[tex]\phi=E2\pi rL[/tex] …… (II)
Here, [tex]E[/tex] is the electric field of charge on cylinder, [tex]L[/tex] is the height or length of curved surface and [tex]r[/tex] is the radius of Gaussian cylindrical curve.
On equating the equation (I) and equation (II) and Rearranging it for [tex]E[/tex] :
[tex]\boxed{E=\dfrac{\lambda}{2\pi\epsilon_0r}}}[/tex]
Thus, the electric field of cylinder at a distance [tex]r[/tex] more than radius of cylinder is equal to [tex]\boxed{E=\dfrac{\lambda}{2\pi\epsilon_0r}}}[/tex].
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Answer Details:
Grade: College
Subject: Physics
Chapter: Electrostatics
Keywords:
Very long, uniformly charged cylinder, radius R, linear charge density λ, cylinder, electric field strength, gauss law, net electric charge and distance r from the axis of the cylinder.
The cylinder's electric field strength outside the cylinder will be "[tex]E = \frac{\lambda }{(2 \pi r)\varepsilon_0 }[/tex]".
According to the question,
The linear charge density of cylinder,
= λ
Let,
As we know,
→ [tex]\oint EdA = \frac{Q}{\varepsilon_0 }[/tex]
or,
→ [tex]E(2 \pi rl) = \frac{\lambda L}{\varepsilon_0 }[/tex]
→ [tex]E = \frac{\lambda L}{(2 \pi rL) \varepsilon_0 }[/tex]
→ [tex]= \frac{\lambda }{(2 \pi r)\varepsilon_0 }[/tex]
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